ENGINEERING COUNCIL
CERTIFICATE LEVEL
ENGINEERING MATERIALS C102
TUTORIAL 5 – NON-MECHANICAL PROPERTIES OF MATERIALS
OUTCOMES
On successful completion of the unit the candidate will be able to:
1.
Recognise the structures of metals, polymers and ceramic materials.
2.
Assess the mechanical and physical properties of engineering materials
3.
Understand the relationships between the structure of a material and its properties.
4.
Select materials for specific engineering applications.
CONTENTS
ELECTRICAL PROPERTIES
• CONDUCTIVITY
• RESISTIVITY
• RELATIVE PERMEABILITY
• RELATIVE PERMITTIVITY
• SEMICONDUCTOR THEORY
THERMAL PROPERTIES
• MELTING POINT
• LATENT HEAT OF FUSION
• THERMAL CONDUCTIVITY
• THERMAL EXPANSION
• TEMPERATURE COEFFICIENT OF RESISTANCE
DURABILITY
• CORROSION RESISTANCE
• OXIDISATION
• CHEMICAL ATTACK
• PERMEABILITY
• DIMENSIONAL STABILITY
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1.
ELECTRICAL PROPERTIES
CONDUCTIVITY
This property governs how well the material conducts electricity. The best material is silver but
this is too expensive so for general use copper is the best compromise. Materials with bad
conductivity may be used as insulators. These are various polymers and ceramics. The formula for
calculating the resistance of a wire is as follows.
R = ρA/L where ρ is the resistivity, L the length and A is the cross sectional area.
RESISTIVITY
The resistance of a conductor increases with length L and decreases with cross sectional area A so
we may say R is directly proportional to L and inversely proportional to A.
R = Constant x L/A
The constant is the resistivity of the material ρ hence R = ρL/A Ohms
SELF ASSESSMENT EXERCISE No.1
1. Calculate the resistance of a copper wire 5 m long and 0.3 mm diameter. The resistivity is 1.7 x
10-8 Ohm metre. (Answer 1.202 Ω)
2. Calculate the resistance of a nichrome wire 2 m long and 0.2 mm diameter given ρ = 108 x 108
(Answer 68.75 Ω)
RELATIVE PERMEABILITY
This is a property that governs the magnetic strength of a material. The symbol is µr. This is a
property that is difficult to give as a constant in tables and you need to understand magnetisation in
depth to follow how to use this property. The main equation that this property occurs is B/H = µoµr
B is the flux density -7in Tesla, H is the magnetising force and µo is the absolute permeability with a
value of 12.566 x 10
RELATIVE PERMITTIVITY
This is a property that governs the electro-static charge stored on an electric capacitor. The symbol
is εr. The main equation that this is found in is C = εoεrA/d
C is the capacitance in Farads and A the area of the capacitor plates and d the distance between
them. εo is the absolute permittivity with a value of εo =8.85 x 10-12
Here is a table of some common figures.
εr
1.000
1.006
2 approx.
7 approx.
4 approx.
6 approx.
various
MATERIAL
Free Space
Air
Paper
Glass
Mica
Ceramic
Plastics
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TEMPERATURE COEFFICIENT OF RESISTANCE
The resistance of most conductors increases with temperature. This is a problem for items like
electrical strain gauges where the changes in resistance must be due to the change in the dimensions
only. (There are special materials like semiconductors where the resistance goes down with
increased temperature covered in the next section). The amount by which the resistance changes per
degree per ohm of the original resistance is called the temperature coefficient of resistance and is
denoted α. The units are Ohms per Ohm per degree.
Let the resistance of a conductor be Ro at 0oC.
Let the resistance be R1 at θ1 oC. The change in resistance = αθ1 Ro
The new resistance is R1 = Ro + αθ1 Ro
Let the resistance be R2 at θ2 oC. The change in resistance = αθ2 Ro
The new resistance is R2 = Ro + αθ2 Ro
If the temperature changes from θ1 to θ2 the resistance changes by
∆R = R2 - R1 = (Ro + αθ2 Ro ) - ( Ro + αθ1 Ro )
∆R = Ro α∆θ
SELF ASSESSMENT EXERCISE No. 2
1. A resistor has a nominal resistance of 120 Ω at 0 oC. Calculate the resistance at 20oC.
Calculate the change in resistance when the temperature drops by 5 degrees. α = 6 x 10 -3 Ω/
Ω oC
(Answers 134.4 Ω and - 3.6Ω)
SEMICONDUCTOR THEORY
There are a group of natural materials that are neither good
conductors nor good insulators. These are called semi
conductors such as Silicon and Germanium. These are used to
make a range of devices that are used in modern electronic
circuits. To understand the electrical properties of these
materials we need to go back to the atomic level. Electrons
orbit in shells of fixed radius representing different fixed
energy bands. The outer band of electrons on an atom is
called the VALENCE band. The cloud of free electrons
surrounding the molecules is called the CONDUCTION
band. In a good conductor the electrons will leave the valence band and join the conduction band
very easily and these electrons are free to form a current when a voltage is applied. The opposite is
true of a good insulator.
The band theory supposes that a fixed amount of energy is required to make an electron jump from
the valence shell into the conduction band. This is called the energy gap. The energy gap is large for
the molecules of a good insulator but for a good conductor it is zero.
The resistance and resistivity of most conductors like copper increases with temperature but in the
case of semi conductors like silicon the resistance goes down. Semi conductors are widely used to
make temperature sensors (e.g. Thermistors) and a simple experiment with one of these will show
that the resistance goes down quite dramatically when plunged into hot water.
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Natural semi conductors are called INTRINSIC. When they are modified by a manufacturing
process to give them enhanced properties, they are then called EXTRINSIC.
The Silicon atom has 14 electrons, two in the inner shell, eight in the second shell, and only four in
the valence shell making it incomplete. This affects the way the atoms bond together in a crystalline
form. There are no free electron and no conduction band. In such materials, the energy required to
make an electron jump the energy gap can come from heat. This means that at temperature above
absolute zero, say room temperature, some electron will jump the energy gap to become free
electrons. The number of electrons in the conduction band rises with temperature and explains the
negative thermal coefficient of resistance.
When an electron makes this jump to the conduction band the parent atom becomes deficient and
most text refers to this as a HOLE because it can be filled by another electron. When current flows
in a semi conductor, the electrons can migrate from atom to atom so these HOLES appear to
migrate in the opposite direction to the electron and this constitutes a current as well. These might
be though of as equivalent to a positively charged particle moving in the opposite direction to the
electron.
Extrinsic semi conductors are produced by doping them to enhance their conductivity. The
conductivity of semiconductors like Silicon can be increased by adding small, controlled amounts
of "impurities" that have roughly the same atomic size, but more or fewer valence electrons than the
semimetal. This process is known as doping.
An impurity with fewer valence electrons such as boron, aluminium or indium takes up space in the
solid structure, but contributes fewer electrons to the valence band, thus generating an electron
deficit and making the atoms more positively charged. This type of doping creates a hole in the
valence band making it possible for the electrons in the valence band to move atom to another
within this band and so increases the conductivity. Such dopes semiconductors are known as p-type
because the atoms are more positively charged.
Alternately, an impurity with more valence electrons such as phosphorus, antimony or arsenic
contributes extra electrons to the band. Since the valence band is already filled by the semimetal,
the extra electrons must go into the conduction band. This also improves the conductivity. Such
dopes semiconductors are known as n-type because the enrichment of electrons makes it more
negatively charged.
SEMICONDUCTOR JUNCTION
Consider what happens when a p-type and n-type material are
brought together to form a junction. On their own there is an
electron surplus in the n-type material and hole surplus in the
p-type. When the two pieces are brought into contact,
electrons from the n-type diffuse into the p-type creating a
junction zone with few charge carriers. This balancing out of
electrons only occurs in the junction zone.
If a voltage is applied across the junction, electrons can easily
move from the p-type to the n-type but cannot flow in the
reverse direction. The junction is a DIODE
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If we use two junctions we have a TRANSISTOR.
You can get more on transistor theory at
http://www.tpub.com/neets/book7/25a.htm
You can have n-p-n or p-n-p transistors. By injecting
or removing electrons at the junction, the junction
zones can be made to conduct and so pass a much
larger current and producing a gain.
LIGHT EMITTING DIODES (LEDs)
These are diodes that give out light when a voltage is applied to them. They are made
with semiconductor materials that have been specially enhanced. When a voltage is
applied to a p-n junction, external electrons from the source flow to the diode and
change the arrangement of electrons in the diode.
Recall that the p-type semiconductor has extra space for electrons in its valence band, and no
electrons in its conduction band, while the n-type semiconductor has a full valence band and extra
electrons in its conduction band.
If the circuit is connected as shown, electrons flow
into the n-type side and occupy the conduction band.
As electrons continue to come into the conduction
band, they will be pushed to the p-type side. The
electrons go into the empty conduction band of the ptype side. Since this is a higher energy band and the
condition is not stable, they then jump to the valence
band making the atoms more stable and in so doing must give up energy as light and the colour
depends on the size of the energy gap. Red is produced by small gaps and yellow by large gaps.
SELF ASSESSMENT EXERCISE No. 3
From the 2004 Exam paper
Explain with reference to band structure and energy levels, the difference between intrinsic and
extrinsic semiconductors.
Temperature has an effect on electrical conductivity. Describe and explain how conductivity
will vary with temperature for the following materials.
Pure Copper
Pure Silicon
Silicon lightly doped with phosphorus.
Describe how semiconductor materials can be used to produce a light-emitting diode (LED).
Describe the operation of the LED. What controls the colour of the light emitted during
operation?
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2.
THERMAL PROPERTIES
MELTING POINT
This is the temperature at which the material melts and it is denoted Tm (K) or θm oC. Here is a
table of the melting points of metals.
Elements
Symbol
θ m oC
Elements
Symbol
θ m oC
Aluminum
Al
659
Manganese
Mn
1260
Brass (85 Cu 15
Zn)
Cu+Zn
900-940
Monel
Ni+Cu+Si
1301
Nickel
Ni
1452
Bronze (90 Cu
10 Sn)
Cu+Sn
850-1000
Phosphorous
P
44
Cast Iron
C+Si+Mn+Fe
1260
Silicon
Si
1420
Carbon
C
3600
Silver
Ag
961
Chromium
Cr
1615
Stainless Steel
Cr+Ni+Mn+C
1363
Copper
Cu
1083
Steel-High
Carbon
Cr+Ni+Mn+C
1353
Gold
Au
1063
Medium Carbon Cr+Ni+Mn+C
1427
Hydrogen
H
-259
Inconel
Ni+Cr+Fe
1393
Iron
Fe
1530
Lead
Pb
327
Magnesium
Mg
Low Carbon
Cr+Ni+Mn+C
1464
Tin
Sn
232
Titanium
Ti
1795
Tungsten
W
3000
Zinc
Zn
419
670
LATENT HEAT OF FUSION
This is the energy required to melt 1 kg of material.
THERMAL CONDUCTIVITY
This is a property that governs how well a material conducts heat. The formula for the heat flow
rate through a wall of area A is given by the following formula.
Φ = k A ∆θ/t
Φ is the heat flow rate in Watts.
k is the thermal conductivity in W/m K
t is the thickness of the wall.
∆θ is the temperature difference between the two sides of the wall.
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THERMAL EXPANSION
When solids and liquids are heated, the molecules vibrate more and take up more space so the
material expands. Consider first the expansion in one direction.
If a bar of material of length Lo has its temperature increased by ∆θ degrees, the increase of length
is ∆L.
This is directly proportional to the original length L and to the temperature change ∆θ. It follows
that :∆L = constant x Lo ∆θ
The constant of proportionality is called the coefficient of linear expansion (α).
∆L = α Lo ∆θ
WORKED EXAMPLE No. 1
A thin steel band 850 mm diameter must be expanded to fit around a disc 851 mm diameter.
Calculate the temperature change needed. The coefficient of linear expansion is 15 x 10-6 per oC
SOLUTION
Initial circumference of ring = πD = π x 850 = 2670.35 mm
Required circumference = π x 851 = 2673.50 mm
∆L = 2673.50 - 2670.35 = 3.15 mm
∆L = α L ∆θ
3.15 = 15 x 10-6 x 2670.35 x ∆θ
∆θ = 3.15/(15 x 10-6 x 2670.35) = 78.6 Kelvin
SUPERFICIAL EXPANSION
This is about the change in area of a flat shape. Consider a flat plate of metal with area Ao. The
change in area is ∆A and this is directly proportional to the temperature change so:∆A = constant x Ao ∆θ
The constant is the coefficient of superficial expansion β
∆A = β Ao ∆θ
Note β = 2α
WORKED EXAMPLE No. 2
A steel sheet has an area of 500 cm2 at 20oC. Calculate the area when it is heated to 300 oC. The
coefficient of superficial expansion is 30 x 10-6 per oC
SOLUTION
∆A = β L ∆θ = 30 x 10-6 x 500 x (300 – 20) = 4.2 cm3
The new area is 504.2 cm2
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CUBICAL EXPANSION
Since a material expands in all direction the volume changes. The change in volume is ∆V.
This is directly proportional to the original volume Vo and to the temperature change ∆θ. It follows
that :∆V = constant x Vo ∆θ
The constant of proportionality is called the coefficient of cubical expansion expansion (γ).
∆L = γ L ∆θ
Note that γ = 2 α
WORKED EXAMPLE No. 3
Calculate the change in volume of 1 m3 of water when it is heated from 10 oC to 80 oC. The
coefficient of cubical expansion is 210 x 10-6 per oC
SOLUTION
∆V = 210 x 10-6 x 1 x (80 -10) = 14.7 x 10-3 m3 or 14.7 dm3 or 14.7 litre
SELF ASSESSMENT EXERCISE No.4
1. A brass bar is 600 mm long and 100 mm diameter. It is heated from 20 oC to 95oC. Calculate
the change in length. α is 18 x 10-6 per oC. (Answer 0.81 mm)
2. A steel ring is 50 mm diameter and 2 mm thick. It must be fitted onto a shaft 50.1 mm diameter.
Calculate the temperature to which it must be heated in order to fit on the shaft. The initial
temperature is 20 oC and the coefficient of linear expansion is 15 x 10-6 per oC.
(Answer 133.3 K)
3. A stub shaft 85.2 mm diameter must be shrunk to 85 mm diameter in order to insert it into a
housing. By how much must the temperature be reduced? Take the coefficient of linear
expansion is 12 x 10-6 per oC.
(Answer -195.6 K)
4. A tank contains 40 m3 of oil at 10oC. Calculate the volume at 40oC given γ = 700 x 10-6 per oC
(0.84 m3)
5. Copper sheet covers a wall and has an area of 20 m2 at 15oC. What is the change in area when
it is heated to 80oC? β = 34 x 10-6 per oC. (44.2 x 10-3 m2)
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3.
DURABILITY
CORROSION RESISTANCE
You will find much information and pictures at http://www.corrosion-doctors.org
OXIDISATION
Corrosion takes many forms and it would require a very large section to explain it. The following is
a brief summary of the forms of corrosion that occur with materials.
Oxygen from the environment combines with the material to form a new substance, usually an
oxide film on the surface. If the oxide film is easily removed to expose new material, the process
will continue until all the material is oxidised. In the case of most ferrous materials the oxide film is
rust and this lets water through and crumbles away. In the case of copper and aluminium, the oxide
film is durable and forms a protective coat on the surface. Materials like Gold and Silver do not
oxidise easily but important engineering materials that resist oxidation are zinc, chromium,
cadmium and others.
ELECTROLYTIC ACTION – BI METAL CORROSION
There are other factors that affect corrosion. The presence of water, especially if it contains salt,
will enable electrolytic action to occur and greatly accelerate the process. This is greatly accelerated
if there are two different metals present to create an electrolytic cell and so care must be taken when
different metals are used in an assembly. The compatibility of metals is listed as the Electrode
Potential (in volts) and the further apart two metals are in the table, the worse the electrolytic
action.
TABLE OF ELECTRODE POTENTIAL IN VOLTS
Gold
+1.68
Lithium
-3.02
Nickel
-0.23
Potassium
-2.92
Tin
-0.14
Sodium
-2.71
Lead
-0.12
Magnesium
-2.34
Hydrogen
0.00
Aluminium
-1.66
Copper
+0.34
Zinc
-0.76
Mercury
+0.8
Iron
-0.44
Silver
+0.8
Cobalt
-0.29
Components and structures are often treated or coated to reduce corrosion. Here is a list of some of
the preparatory treatments and coatings used for metals.
•
•
•
•
•
•
•
•
PROTECTIVE COATINGS
• Galvanising
• Sherardising
• Calorising
• Chromising
• Chromating
• Phosphating
• Metal Spraying
• Cladding
• Anodising
• Electroplating
• Plastic Coating
• Paint Coating
TREATMENT
Pickling
Degreasing.
Wire brushing
Shot and vapour blasting
Flame descaling
Abrasive finishing
Polishing
Barrelling
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CHEMICAL ATTACK
Metals may be degraded by a variety of chemicals such as acids and alkali and these must always
be considered.
Polymers do not corrode but may be prone to attack from other chemicals such as SOLVENTS
which dissolves them away.
Ceramics are in the main resistant to most forms of chemicals and this is why they have been used
for containers down the centuries.
POROSITY
We should mention here the importance of porosity. If a material allows liquid or gas to seep
through it then this may cause problems such as contamination of food stuffs. On the other hand,
porosity is necessary for a material used as a filter.
DIMENSIONAL STABILITY
We should also mention dimensional stability here. A material subjected to prolonged heat, cold,
pressure and stress may change its dimensions and shape (e.g. creep covered in a later tutorial).
Plastic bottles containing pressurised fizzy drinks should not change their shape over its intended
shelf life.
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